Question

how many term of the ap 9, 17 ,25 must be taken that their sum is 636

Answer 1

a=9
common difference d=8
let sum of n terms be 636
$636 = \frac{x}{2} (2 \times 9+ (x - 1)8) \\ 636= \frac{x}{2 } (18 + 8x - 8) \\ 636 = \frac{x}{2} (10 + 8x) \\ 636 = \frac{x \times 2}{2} (5 + 4x) \\ 636 = 5x + 4 {x}^{2} \\ 4 {x}^{2} + 5 x- 636 = 0 \\ 4 {x}^{2} + 53x - 48x - 636 = 0 \\ x(4x + 53) - 12(4x + 53) = 0 \\ (4x + 53)(x - 12) = 0 \\ x = 12$
hence sim of first 12 terms will 636

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let: first\: term\; be\: a \:and\: CD\: = 17 - 9 = 8$

$\bf\huge => S_{n} = 636$

$\bf\huge => \frac{N}{2}[2a + (n - 1)d] = 636$

$\bf\huge => \frac{N}{2}[2\times 9 + (n - 1)8] = 636$

$\bf\huge => \frac{N}{2} (8n - 10) = 636$

$\bf\huge => n(4n + 5) = 636$

$\bf\huge => 4n^2 + 5n + 636 = 0$

$\bf\huge => n = \frac{-5 + \sqrt{25 - 4\times 4\times -636}}{2\times 4}$

$\bf\huge = \frac{-5 + \sqrt{25 + 10176}}{8}$

$\bf\huge = \frac{- 5 + \sqrt{10201}}{8}$

$\bf\huge = \frac{-5 + 101}{8}$

$\bf\huge = \frac{96}{8} , \frac{-106}{8}$

$\bf\huge = 12 , \frac{-53}{4}$

$\bf\huge But\: n \:cannot\: be\: Negative$

$\bf\huge => n = 12$

$\bf\huge Hence\:Sum\: of\: 12\: terms\: is\: 636$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$